The calculation of the wall stiffnesses refers to the second article of this series.
1
Building
Floor Plan
Due to its different stiffness, the wall stiffness has a considerable influence on the total deformation of the floor plan. Furthermore, the asymmetrical arrangement of the walls also influences the deformation of the building.
Usually, this effect is taken into account by stiffnesses of a 2D position.
Example
This effect is demonstrated on a simple two-story building. The building has a regular floor plan. Further information on minimum stiffening criteria can be found in [1].
1
Building
System
- Ground plan = 5 m ⋅ 10 m
- Simplified calculation, windows are excluded story-high.
- A tie rod is placed on each end of the wall.
- Wall structure and stiffnesses as described in article two of this series.
- FE mesh size = 1.5 m
Loading
- Self-weight and structure = 2 kN/m²
- Wind load in global y-direction
- WZ2
- Height = 100 m above sea level NN
- we+d = 0.46 + 0.74 = 1.2 kN/m²
- We,d = 1.2 kN/m² ⋅ 3 m = 3.6 kN/m
Combination
- CO1 = 1.0 LC1 + 1.5 LC2
Wall Stiffnesses
This results in four different wall lengths. To simplify it, the wall stiffnesses are calculated for equivalent members. The determination of the stiffnesses is the samle like in the previous article.
| Shear Wall | Length [m] | Modulus of Elasticity [kN/cm²] | D66/D77 [kN/cm] | G-Modulus [kN/cm²] | D88 [kN/cm] | Stiffness Support [kNcm/rad] |
|---|---|---|---|---|---|---|
| 1 | 0.5 | 792 | 9,504 | 0.47 | 6.5 | 64,499 |
| 2 | 1.0 | 396 | 4,752 | 0.80 | 11.0 | 257,995 |
| 3 | 1.5 | 264 | 3,168 | 1.04 | 14.3 | 580,489 |
| 4 | 2.5 | 158 | 1,901 | 1.36 | 18.8 | 1.612.469 |
The wall stiffnesses are calculated for each of these four wall lengths. For this purpose, each wall is loaded with a unit load of 1 kN. Since wall lengths over 2.5 m with a height of 2.75 m are not possible to manufacture, the middle wall is divided in the middle.
In the attached RFEM model file 1, the deformations for all wall lengths are calculated as surfaces and member results. In the upper part of the model, the deformation is calculated without a wall anchor and in the lower part with a wall anchor. The deformations are also compared in Image 04.
From the determined deformations of the individual walls, a stiffness is calculated for each wall.
As an example, the following stiffness is obtained for wall 1 with a length of 50 cm:
C = F /u = 1 kN / 22.5 mm = 0.044 kN/mm
c = F / l ⋅ C = 1 kN / 0.5 m ⋅ 0.044 kN/mm = 0.088 N/mm²
For all walls:
- Wall | l = 0.5 m | c = 0.088 N/mm²
- Wall | l = 1.0 m | c = 0.164 N/mm²
- Wall | l = 1.5 m | c = 0.230 N/mm²
- Wall | l = 2.5 m | c = 0.333 N/mm²
These stiffnesses are assigned to the respective line support in the floor plan (see Image 05). The floor plan can be found in the attached RFEM model file 2.
Due to the symmetrical building, there is no rotation of the building. This article explains this in more detail:
The attached video shows how the horizontal forces develop in an asymmetrical floor plan against the load direction.
Conclusion
This article showed the calculation by floor of timber panel buildings. The stiffnesses can be determined by means of surface or member elements. Elasticities resulting from an anchorage are taken into account.
Use the attached Excel file to reproduce the calculation of the examples.
A final part of this series will follow, which shows the design of the forces determined from the line support reactions (see Image 05).
Knowledge Base